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Lời giải:
a) \(\frac{45x(3-x)}{15(x-3)^3}=\frac{-45x(x-3)}{15(x-3)^3}=\frac{-3x}{(x-3)^2}\)
b) \(\frac{36(x-2)^3}{32-16x}=\frac{36(x-2)^3}{-16(x-2)}=\frac{-9}{4}(x-2)^2\)
c) \(\frac{x^2-xy}{5y^2-5xy}=\frac{x(x-y)}{-5y(x-y)}=\frac{x}{-5y}\)
d) \(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{-(x^2-y^2)}{(x-y)^3}=\frac{-(x-y)(x+y)}{(x-y)^3}=\frac{-(x+y)}{(x-y)^2}\)
a) Tìm MTC: x3 – 1 = (x – 1)(x2 + x + 1)
Nên MTC = (x – 1)(x2 + x + 1)
Nhân tử phụ:
(x3 – 1) : (x3 – 1) = 1
(x – 1)(x2 + x + 1) : (x2 + x + 1) = x – 1
(x – 1)(x2+ x + 1) : 1 = (x – 1)(x2 + x + 1)
Qui đồng:
b) Tìm MTC: x + 2
2x – 4 = 2(x – 2)
6 – 3x = 3(2 – x)
MTC = 6(x – 2)(x + 2)
Nhân tử phụ:
6(x – 2)(x + 2) : (x + 2) = 6(x – 2)
6(x – 2)(x + 2) : 2(x – 2) = 3(x + 2)
6(x – 2)(x + 2) : -3(x – 2) = -2(x + 2)
Qui đồng:
\(a,\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)
\(b,\dfrac{2y}{2x^2-xy}+\dfrac{4x}{xy-2x^2}\)
\(=\dfrac{2y}{2x^2-xy}-\dfrac{4x}{2x^2-xy}\)
\(=\dfrac{2y-4x}{2x^2-xy}\)
\(=\dfrac{2\left(y-2x\right)}{x\left(2x-y\right)}\)
\(=\dfrac{-2\left(2x-y\right)}{x\left(2x-y\right)}\)
\(=-\dfrac{2}{x}\)
\(c,\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2}{x-3}-\dfrac{2x^2-2x}{x-3}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x^2+2x+5-4x}{x-3}\)
\(=\dfrac{-3x^2-2x+9}{x-3}\)
\(a,\dfrac{16+x}{x^2-2x}+\dfrac{18}{2x-x^2}\)
\(=\dfrac{16+x}{x^2-2x}-\dfrac{18}{x^2-2x}\)
\(=\dfrac{16+x-18}{x^2-2x}\)
\(=\dfrac{x-2}{x\left(x-2\right)}\)
\(=\dfrac{1}{x}\)
\(b,\dfrac{2y}{2x^2-xy}+\dfrac{4x}{xy-2x^2}\)
\(=\dfrac{2y}{2x^2-xy}-\dfrac{4x}{2x^2-xy}\)
\(=\dfrac{2y-4x}{2x^2-xy}\)
\(=\dfrac{2\left(y-2x\right)}{x\left(2x-y\right)}\)
\(=\dfrac{-2\left(2x-y\right)}{x\left(2x-y\right)}\)
\(=-\dfrac{2}{x}\)
\(c,\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2}{x-3}-\dfrac{2x^2-2x}{x-3}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x^2+2x+5-4x}{x-3}\)
\(=\dfrac{-3x^2-2x+9}{x-3}\)
a) −x2+21−5x−x2+21−5x =x2+2−(1−5x)=x2+2−(1−5x) =x2+25x−1=x2+25x−1;
b) −4x+15−x−4x+15−x =4x+1−(5−x)=4x+1−(5−x) =4x+1x−5
\(a,\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)
\(=\frac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\)
\(=\frac{2\left(x-2\right)}{x+2}\)
Với \(x=\frac{1}{2}\)
\(\Rightarrow\frac{2\left(x-2\right)}{x+2}=\frac{2\left(\frac{1}{2}-2\right)}{\frac{1}{2}+2}=\frac{2.-\frac{3}{2}}{\frac{5}{2}}=-3.\frac{2}{5}=\frac{-6}{5}\)
b,Do x = -5; y = 10=> y = -2x
Thay y = -2x vào biểu thức ta được
\(\frac{x^3-x^2\left(-2x\right)+x\left(-2x\right)^2}{x^3+\left(-2x\right)^3}\)
\(=\frac{x^3+2x^3+2x^2}{x^3-8x^3}\)
\(=\frac{3x^3+2x^2}{-7x^3}=\frac{3}{-7}+\frac{2}{-7x}\)
Thay x = -5 là đc
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
a) \(\dfrac{36\left(x-2\right)^3}{32-16x}=\dfrac{36\left(x-2\right)^3}{16\left(2-x\right)}=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}\)\(=\dfrac{36\left(x-2\right)^3:4\left(x-2\right)}{-16\left(x-2\right):4\left(x-2\right)}\)\(=\dfrac{9\left(x-2\right)^2}{-4}\)
b) \(\dfrac{x^2-xy}{5y^2-5xy}=\dfrac{x\left(x-y\right)}{5y\left(y-x\right)}=\dfrac{x\left(x-y\right)}{-5y\left(x-y\right)}\)\(=\dfrac{x}{-5y}\)